A 230 g air-track glider is attached to a spring. The glider is pushed in 8.2 cm and released. A student with a stopwatch finds that 13 oscillations take 19.0 s . What is the spring constant? Express your answer using two significant figures.

Answer:

(a) 9.36 kHz

(b) 3.12 kHz

Explanation:

(a)

V = speed of sound

[tex]v[/tex] = speed of airplane = (0.5) V

f = actual frequency of sound emitted by airplane = 4.68 kHz = 4680 Hz

f' = Frequency heard by the stationary listener

Using Doppler's effect

[tex]f' = \frac{Vf}{V-v}[/tex]

[tex]f' = \frac{V(4680)}{V-(0.5)V)}[/tex]

f' = 9360 Hz

f' = 9.36 kHz

(b)

V = speed of sound

[tex]v[/tex] = speed of airplane = (0.5) V

f = actual frequency of sound emitted by airplane = 4.68 kHz = 4680 Hz

f' = Frequency heard by the stationary listener

Using Doppler's effect

[tex]f' = \frac{Vf}{V+v}[/tex]

[tex]f' = \frac{V(4680)}{V+(0.5)V)}[/tex]

f' = 3120 Hz

f' = 3.12 kHz

The frequencies heard by the stationary listener when the airplane is approaching and when it is moving away can be calculated using the formula for the Doppler effect. The formula differs slightly depending on whether the source of the sound (in this case, the airplane) is moving towards or away from the observer (the listener).

This question involves the Doppler Effect, which describes how the frequency of a wave changes for an observer moving relative to the source of the wave. We are given that the airplane is moving at half the speed of sound and emits a sound of frequency 4.68 kHz.

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Answer:

[tex]U = 1.83 \times 10^5 J[/tex]

Explanation:

Total internal energy of the gas is given by the formula

[tex]U = \frac{f}{2}nRT[/tex]

here as we know that neon gas is monoatomic gas

so we will have

[tex]f = 3[/tex]

n = 50 moles

[tex]T = 20 ^0C = 20 + 273 = 293 K[/tex]

now from above equation we will have

[tex]U = \frac{3}{2}(50)(8.31)(293)[/tex]

[tex]U = 1.83 \times 10^5 J[/tex]

Answer:

Part a)

[tex]\lambda = 300 m[/tex]

Part b)

[tex]E = 2.7 N/C[/tex]

Part c)

[tex]I = 9.68 \times 10^{-3} W/m^2[/tex]

[tex]P = 3.22 \times 10^{-11} N/m^2[/tex]

Explanation:

Part a)

As we know that frequency = 1 MHz

speed of electromagnetic wave is same as speed of light

So the wavelength is given as

[tex]\lambda = \frac{c}{f}[/tex]

[tex]\lambda = \frac{3\times 10^8}{1\times 10^6}[/tex]

[tex]\lambda = 300 m[/tex]

Part b)

As we know the relation between electric field and magnetic field

[tex]E = Bc[/tex]

[tex]E = (9 \times 10^{-9})(3\times 10^8)[/tex]

[tex]E = 2.7 N/C[/tex]

Part c)

Intensity of wave is given as

[tex]I = \frac{1}{2}\epsilon_0E^2c[/tex]

[tex]I = \frac{1}{2}(8.85 \times 10^{-12})(2.7)^2(3\times 10^8)[/tex]

[tex]I = 9.68 \times 10^{-3} W/m^2[/tex]

Pressure is defined as ratio of intensity and speed

[tex]P = \frac{I}{c} = \frac{9.68\times 10^{-3}}{3\times 10^8}[/tex]

[tex]P = 3.22 \times 10^{-11} N/m^2[/tex]

Answer:

a) Travel time = 60.56 s

b) Maximum speed = 36.33 m/s

Explanation:

a)  Distance = 1.1 km = 1100 m

 A subway train accelerates at +1.2 m/s² from rest  through the first half of the distance and decelerates at −1.2 m/s² through the second half.

 So half the distance is traveled at an acceleration of +1.2 m/s².

 We have equation of motion [tex]s=ut+\frac{1}{2}at^2[/tex]

 Substituting

         [tex]\frac{1100}{2}=0\times t+\frac{1}{2}\times 1.2t^2\\\\t=30.28s[/tex]

 Travel time = 2 x 30.28 = 60.56 s

b) We have equation of motion v = u+at

   Substituting t = 30.28 s and a = 1.2 m/s²

   v = 0 + 1.2 x 30.28 = 36.33 m/s

   Maximum speed = 36.33 m/s

c) Photos of graphs are given

Answer:

a) ttotal=60.58 s

b) v^2=36.3 m/s

c) the attached photo shows the graphics.

Explanation:

a) we have the following:

S1=1.1x10^-3/2=550 m

a=1.2 m/s

we have the following formula:

v^2=u^2+2as

replacing values:

v^2=0+(2*1.2*550)=36.33 m/s

The time is equal to:

t1=v/a=36.33/1.2=30.3 s

we have the following:

u=36.33 m/s

v=0

S=550 m/s

using the following expression:

v=u+at2

clearing t2:

t2=(v-u)/a=(0-36.33)/-1.2=30.28 s

ttotal=t1+t2=30.3+30.28=60.58 s

b) the mass speed is equal to:

v^2=u^2+2as=36.3 m/s

Answer:

0.24 kgm²

Explanation:

[tex]L[/tex] = length of the bat = 81.3 cm = 0.813 m

[tex]m[/tex]  = mass of the bat = 0.96 kg

[tex]d[/tex]  = distance of the center of mass of bat from the axis of rotation = 55.9 cm = 0.559 m

[tex]T[/tex]  = Period of oscillation = 1.35 sec

[tex]I[/tex] = moment of inertia of the bat

Period of oscillation is given as

[tex]T = 2\pi \sqrt{\frac{I}{mgd}}[/tex]

[tex]1.35 = 2(3.14) \sqrt{\frac{I}{(0.96)(9.8)(0.559)}}[/tex]

[tex]I[/tex] = 0.24 kgm²

Answer:

The resistivity of the material used to make the rod is ρ= 7.5 * 10⁻⁷ Ω.m

Explanation:

R= 0.2 Ω

L= 0.8 m

S= 1.5mm*2mm= 3 mm² = 3 * 10⁻⁶ m²

ρ = (R*S)/L

ρ= 7.5 * 10⁻⁷ Ω.m

Answer:

Loss in kinetic energy is 86.4 J.

Explanation:

It is given that,

Mass of first car, m₁ = 3 kg

Velocity of first car, v₁ = 9 m/s

Mass of second car, m₂ = 2 kg

Velocity of second car, v₂ = -3 m/s (opposite direction)

If the cars are locked together after the collision with a speed of 4.20 m/s, V = 4.2 m/s

It is a case of inelastic collision. Some of the kinetic energy will be lost in the form of heat energy, sound energy etc.

Initial kinetic energy, [tex]K_i=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2[/tex]

[tex]K_i=\dfrac{1}{2}\times 3\ kg\times (9\ m/s)^2+\dfrac{1}{2}\times 2\ kg\times (-3\ m/s)^2[/tex]

[tex]K_i=130.5\ J[/tex]

Final kinetic energy, [tex]K_f=\dfrac{1}{2}(m_1+m_2)V^2[/tex]

[tex]K_f=\dfrac{1}{2}\times (3\ kg+2\ kg)\times (4.2\ m/s)^2[/tex]

[tex]K_f=44.1\ J[/tex]

Kinetic energy lost, [tex]\Delta K=K_f-K_i[/tex]

[tex]\Delta K=44.1-130.5[/tex]

[tex]\Delta K=-86.4\ J[/tex]

So, 86.4 J of kinetic energy is lost. Hence, this is the required solution.

The loss in kinetic energy during a collision can be calculated by computing the initial and final kinetic energies of the objects and then finding the difference.

In order to compute the loss in kinetic energy, we first need to compute the initial and final kinetic energies of both toy cars. Initially, the kinetic energy of the 3 kg toy car is (1/2) * 3 kg * (9 m/s)^2 and that of the 2 kg toy car is (1/2) * 2 kg * (3 m/s)^2. Summing these, we obtain the initial kinetic energy (call this Ki). Following the collision, the toy cars stick together and move with a common velocity of 4.20 m/s. Therefore, the final kinetic energy (call this Kf) of them together is (1/2) *  (3 kg + 2 kg) * (4.20 m/s)^2. The loss in kinetic energy is then Ki - Kf.

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Answer:

23.99N

Explanation:

Given:

Pressure created by the heart = 120mm of hg

converting the pressure into the standard unit of N/m²

1mm of hg = [tex]\frac{1}{760}atm=\frac{1}{760}\times 1.013\times10^5N/m^2[/tex]

now, 120 mm of hg in N/m² will be

120mm of hg = [tex]\frac{12}{760}\times 1.013\times10^5N/m^2[/tex]

also

given effective area = 15.0 cm² = 15 × 10⁻⁴m²

Now,

Force = Pressure × Area

thus,

Force exerted will be =  [tex]\frac{120}{760}\times 1.013\times10^5N/m^2[/tex] ×  15 × 10⁻⁴m²

or

Force exerted will be = 23.99N

The force that it exerts is about 24.0 N

[tex]\texttt{ }[/tex]

Let's recall Hydrostatic Pressure formula as follows:

[tex]\boxed{ P = \rho g h}[/tex]

where:

P = hydrosatic pressure ( Pa )

ρ = density of  fluid ( kg/m³ )

g = gravitational acceleration ( m/s² )

h = height of a column of liquid ( m )

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

blood pressure = P = 120 mmHg = 0.12 mHg

effective area = A = 15.0 cm² = 15.0 × 10⁻⁴ m²

density of mercury = ρ = 13600 kg/m³

gravitational acceleration = g = 9.8 m/s²

Asked:

force = ?

Solution:

We will use this following formula to solve this problem:

[tex]P = F \div A[/tex]

[tex]F = P A[/tex]

[tex]F = \rho g h A[/tex]

[tex]F = 13600 \times 9.8 \times 0.12 \times ( 15.0 \times 10^{-4} )[/tex]

[tex]F \approx 24.0 \texttt{ N}[/tex]

[tex]\texttt{ }[/tex]

The force that it exerts is about 24.0 N

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Pressure

Answer:

increase the charge

Explanation:

The force acting on charge particle when moving perpendicular to the magnetic field

F = q v B

The centripetal force is given by

F =  mv^2 / r

Comparing both, we get

r = m v / B q

That means the radius of the circular path depends on mass of the charge particle, velocity of the charge particle, magnetic field strength and charge of the particle.

To decrease the radius:

1. increase the charge

2. increase the magnetic field strength

3. decrease the speed

4. decrease the mass

The factor that will decrease the radius of the circular path is increase the charge on the particle.

The magnetic force on the charged particle is calculated as follows;

F = qvB

Where;

The force on the particle moving in circular path is given as;

F = mv²/r

qvB = mv²/r

qB = mv/r

r = mv/qB

Thus, the factor that will decrease the radius of the circular path is increase the charge on the particle.

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Answer:

0.3659

Explanation:

The power (p) is given as:

P = AeσT⁴

where,

A =Area

e = transmittivity

σ = Stefan-boltzmann constant

T = Temperature

since both the bulbs radiate same power

P₁ = P₂

Where, 1 denotes the bulb 1

2 denotes the bulb 2

thus,

A₁e₁σT₁⁴ = A₂e₂σT₂⁴

Now e₁=e₂

⇒A₁T₁⁴ = A₂T₂⁴

or

[tex]\frac{A_1}{A_2} =\frac{T_{2}^{4}}{T_{1}^{4}}[/tex]

substituting the values in the above question we get

[tex]\frac{A_1}{A_2} =\frac{2100_{2}^{4}}{2700_{1}^{4}}[/tex]

or

[tex]\frac{A_1}{A_2} }[/tex]=0.3659

At the frequency of 5 MHz, the period of the oscillations is 1/5meg. That's a period of 1/5 microsecond.

There are 5 full cycles in one full microsecond, and there are 2.5 full cycles in a 0.5 us pulse.

You'll have to decide for yourself how damped a pulse of 2.5 cycles is, because the parameters of the definition are corrupted in the question.

The induced current in a loop inside a sinusoidally-driven solenoid can be determined from Faraday's law of electromagnetic induction. The time derivative of the changing flux through the loop, when divided by the resistance of the loop, provides the value of the induced current.

The question asks about the current induced in a loop placed inside a solenoid driven by an alternating current. This is a physics problem related to electromagnetic induction. The alternating magnetic field in the solenoid induces an electric field that in turn induces a current in the loop, by Faraday's Law.

In this case, the magnetic field B inside the solenoid is oscillating sinusoidally with time as B(t) = B cos(ωt). Thus, the magnetic flux Φ through the loop is changing with time. Using Faraday's law, this flux change induces an emf in the loop, which induces a current I. The emf -dΦ/dt is equal to I*R where R is the resistance of the loop. Hence the induced current I as a function of time can be given by I = -(1/R) * (d/dt) (B cos(ωt) * π * (a/2)^2).

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Answer:

There is a dependency relationship between the refractive index of each substance and the radiation wavelength.

The refractive index in a given medium is inversely proportional to the wavelength of a color.

For example:

The rays of the red color have a wavelength greater than the rays of the blue color, therefore they have a lower refractive index and consequently a light scattering less than the blue.

Snell's law :

n₂/n₁ = v₁/v₂ =  λ₁ /λ₂

*n: (refractive index)

v: (speed of light propagation)

λ: (wavelength)

Answer:

ZERO

Explanation:

As we know that force due to magnetic field on moving charge is given by

[tex]\vec F = q(\vec v \times \vec B)[/tex]

now here we can say that the force due to magnetic field is always perpendicular to the velocity and it is also perpendicular to the magnetic field.

So we can say that

[tex]\vec F. \vec v = 0[/tex]

so power due to magnetic field is always zero

which shows that rate of work done by magnetic field on moving charge is zero

so here the work done by magnetic field is always zero

Answer:

E

Explanation:

F = G * m1 * m2 / r^2     Increase the distance by 3

F1 = G * m1 * m2 / (3r)^2

F1 = G * m1 * m2 / (9*r^2) What this means is the the force decreases by a factor of 9, but we are not done.

F2 = G * 3m1 * 3m2 / (9 r^2)

F2 = G * 9 m1 * m2 / (9 r^2)

In F2 the 9s cancel out and we are left with

F2 = G * m1 * m2/r^2 which is the same thing.

F2 equals F

Answer:

E. The gravitational force remains unchanged

Explanation:

(a) [tex]50.4 N\cdot m[/tex]

The torque exerted on the solid disk is given by

[tex]\tau=Frsin \theta[/tex]

where

F is the magnitude of the force

r is the radius of the disk

[tex]\theta[/tex] is the angle between F and r

Here we have

F = 180 N

r = 0.280 m

[tex]\theta=90^{\circ}[/tex] (because the force is applied tangentially to the disk)

So the torque is

[tex]\tau = (180 N)(0.280 m)(sin 90^{\circ})=50.4 N\cdot m[/tex]

(b) [tex]17.2 rad/s^2[/tex]

First of all, we need to calculate the moment of inertia of the disk, which is given by

[tex]I=\frac{1}{2}mr^2[/tex]

where

m = 75.0 kg is the mass of the disk

r = 0.280 m is the radius

Substituting,

[tex]I=\frac{1}{2}(75.0)(0.280)^2=2.94 kg m^2[/tex]

And the angular acceleration can be found by using the equivalent of Newtons' second law for rotational motions:

[tex]\tau = I \alpha[/tex]

where

[tex]\tau = 50.4 N \cdot m[/tex] is the torque exerted

I is the moment of inertia

[tex]\alpha[/tex] is the angular acceleration

Solving for [tex]\alpha[/tex], we find:

[tex]\alpha = \frac{\tau}{I}=\frac{50.4 N \cdot m}{2.94 kg m^2}=17.2 rad/s^2[/tex]

The resulting velocity of the tennis ball is calculated by first determining the impulse, which is the integral of the force over the contact time. The impulse is then equal to the change in momentum, allowing us to solve for the velocity. The calculated velocity of the ball after impact is 74.67 m/s.

To calculate the resulting velocity of the ball after being hit by the racket, we first need to determine the impulse imparted to the ball. The force applied by the racket is variable and given by F=at-bt², where a = 1290 N/ms, b = 330 N/ms², and t is the time in milliseconds. To calculate the impulse (J), we integrate the force over the contact time, from 0 to 2.80 ms.

Impulse, J, is the integral of F with respect to t, which gives us J = (1/2)at² - (1/3)bt³ evaluated from 0 to 2.80 ms. Plugging in the values:

J = (1/2)(1290 N/ms)(2.80 ms)² - (1/3)(330 N/ms²)(2.80 ms)³ = 1290(3.92) - 330(2.744) N = 5056.8 - 905.52 N = 4151.28 Nms

The impulse is equal to the change in momentum of the tennis ball. Considering the ball's mass m = 55.6 g = 0.0556 kg, and initial velocity, u = 0 m/s (since it's hit from rest), we apply the impulse-momentum theorem:

J = F - mv, therefore, v = J/m.

Substituting the values we have: v = 4151.28 Nms / 0.0556 kg = 74667.7 m/s = 74.67 m/s.

So, the resulting velocity of the tennis ball is 74.67 m/s after the racket's impact.

The resulting velocity of the ball after being hit by the racket is approximately [tex]\( 49466.325 \, \text{m/s} \)[/tex].

To find the resulting velocity of the ball after being hit by the racket, we can use the impulse-momentum theorem, which states that the change in momentum of an object is equal to the impulse applied to it:

[tex]\[ J = \Delta p \][/tex]

The impulse J is equal to the integral of the force F with respect to time t over the duration of the contact:

[tex]\[ J = \int_{0}^{2.80} F \, dt \][/tex]

Given that [tex]\( F = at - bt^2 \)[/tex], we can integrate [tex]\( F \)[/tex] with respect to [tex]\( t \)[/tex] to find [tex]\( J \)[/tex]:

[tex]\[ J = \int_{0}^{2.80} (at - bt^2) \, dt \][/tex]

[tex]\[ J = \left[ \frac{1}{2} at^2 - \frac{1}{3} bt^3 \right]_{0}^{2.80} \][/tex]

[tex]\[ J = \left( \frac{1}{2} a(2.80)^2 - \frac{1}{3} b(2.80)^3 \right) - \left( \frac{1}{2} a(0)^2 - \frac{1}{3} b(0)^3 \right) \][/tex]

[tex]\[ J = \left( \frac{1}{2} \cdot 1290 \cdot (2.80)^2 - \frac{1}{3} \cdot 330 \cdot (2.80)^3 \right) - 0 \][/tex]

[tex]\[ J = \left( \frac{1}{2} \cdot 1290 \cdot 7.84 - \frac{1}{3} \cdot 330 \cdot 21.952 \right) \][/tex]

[tex]\[ J = \left( 5056.64 - 2308.656 \right) \][/tex]

[tex]\[ J = 2747.984 \, \text{N} \cdot \text{ms} \][/tex]

Now, we know that impulse [tex]\( J \)[/tex] is also equal to the change in momentum [tex]\( \Delta p \)[/tex]  of the ball:

[tex]\[ J = \Delta p = mv - mu \][/tex]

Where:

- [tex]\( m \)[/tex] is the mass of the ball,

- v is the final velocity of the ball,

- u is the initial velocity of the ball (which we assume to be zero as the ball is initially at rest).

So, we can rearrange the equation to solve for v:

[tex]\[ v = \frac{J}{m} \][/tex]

Now, let's plug in the values to find v:

[tex]\[ v = \frac{2747.984 \, \text{N} \cdot \text{ms}}{0.0556 \, \text{kg}} \][/tex]

[tex]\[ v \approx 49466.325 \, \text{m/s} \][/tex]

So, the resulting velocity of the ball after being hit by the racket is approximately [tex]\( 49466.325 \, \text{m/s} \)[/tex].

Answer:

100 Degrees is boiling point.

Explanation:

Explanation:

When a body or object falls, basically two forces act on it:  

1. The force of air friction, also called "drag force" [tex]D[/tex]:  

[tex]D={C}_{d}\frac{\rho V^{2} }{2}A[/tex]  (1)

Where:  

[tex]C_ {d}[/tex] is the drag coefficient  

[tex]\rho[/tex] is the density  of the fluid (air for example)

[tex]V[/tex] is the velocity  

[tex]A[/tex] is the transversal area of the object

So, this force is proportional to the transversal area of ​​the falling element and to the square of the velocity.  

2. Its weight due to the gravity force [tex]W[/tex]:  

[tex]W=m.g[/tex]

(2)

Where:  

[tex]m[/tex] is the mass of the object

[tex]g[/tex] is the acceleration due gravity  

So, at the moment when the drag force equals the gravity force, the object will have its terminal velocity:

[tex]D=W[/tex] (3)

[tex]{C}_{d}\frac{\rho V^{2} }{2}A=m.g[/tex]  (4)

[tex]V=\sqrt{\frac{2m.g}{\rho A{C}_{d}}}[/tex]  (5) This is the terminal velocity

As we can see, there is no "falling time" in this equation.

Therefore, the terminal velocity is not dependent on the falling time.

Answer:

Velocity of golf ball≅ 12.87mph    

Explanation:

Using the theory of conservative momentum to both the club head and the golf ball  we have;

Qi1 + Qi2= Qf1 + Qf2

Qi1: initial momentum for the club head

Qf1: final momentum for the club head

Qi2: initial momentum for the golf ball

Qf2: final momentum for the golf ball

momentum (Q) = Mass x velocity

which means the sum of the momentums of both club head and golf ball has to be the same before and after they have collided.

using the Coeficient of restitution e= 0.83 allows us to know what kind of collision we are dealing with, which is a partially elastic collision since

0> e=0.83 >1.

Qi1= 190 x 94= 17,860 mph.g

Qf1= 190 x Vf1

Qi2= 46 x 0= 0 mph.g

Qf2= 46x Vf2

Using the value of e to determine Vf2 as the final velocity of the golf ball:

e= [tex]\frac{Vf2- Vf1}{Vi1 - Vi2}[/tex]

0.83= [tex]\frac{Vf2 - Vf1}{94 - 0}[/tex]

Vf1= (0.83 x 94)+ Vf2

Vf1= 78.02 + Vf2

Qi1 + Qi2= Qf1 + Qf2

17,860 + 0 = 190xVf1 + 46xVf2    

17,860= 190x (78.02+Vf2) + 46xVf2

17,860= 14,823+ 190xVf2+46xVf2

Vf2≅ 12.87mph                

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

[tex]Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}[/tex]

Where,

[tex]C_d[/tex] is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = [tex]\frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2[/tex]

A₂ = Area at the throat

A₂ = [tex]\frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2[/tex]

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

[tex]Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}[/tex]

or

[tex]Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}[/tex]

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Answer:

Part a)

[tex]T = 425.6 K[/tex]

Part b)

[tex]KE_{avg} = 8.81\times 10^{-21} J[/tex]

Part c)

in order to find the average speed we need to know about the the gas molar mass or we need to know which gas it is.

Explanation:

Part a)

As per ideal gas equation we know that

[tex]PV = nRT[/tex]

here we know that

[tex]P = 1.60 \times 10^6 Pa[/tex]

n = 3.80 moles

[tex]V = 8.40 L = 8.40 \times 10^{-3} m^3[/tex]

now from above equation we have

[tex]T = \frac{PV}{nR}[/tex]

[tex]T = \frac{(1.60 \times 10^6)(8.40 \times 10^{-3})}{(8.31)(3.80)}[/tex]

[tex]T = 425.6 K[/tex]

Part b)

Average kinetic energy of the gas is given as

[tex]KE_{avg} = \frac{3}{2}KT[/tex]

here we know that

[tex]K = 1.38 \times 10^{-23}[/tex]

T = 425.6 K

now we have

[tex]KE_{avg} = \frac{3}{2}(1.38 \times 10^{-23})(425.6)[/tex]

[tex]KE_{avg} = 8.81\times 10^{-21} J[/tex]

Part c)

in order to find the average speed we need to know about the the gas molar mass or we need to know which gas it is.

Answer:

W = 627.2 J

Explanation:

Given:

[tex]\rho_{chain}[/tex] = 2kg/m

length of chain = 12 m

length pulled will be = 8 m

We know

Work done (W) = mgh

where

m = mass of the object

g= acceleration due to gravity

h = displacement

For a small length dy of the chain, the work done can be written as:

dW  = (mass of the small length pulled)× g×dy

dW = 2kg/m ×dy×9.8×y

where, y is the distance from the ground level of the end of chain

integerating the above equation

W = [tex]\int\limits^8_0 {19.6y} \, dx[/tex]

W =[tex][19.6\frac{y^2}{2}]_{0}^{8}[/tex]

W = 627.2 J

Answer:

55536.6 J

Explanation:

Given:

Mass of the ice = 54g

Initial temperature = 0°C

Final Temperature = 100°C

Mass of the steam = 6.6g

Now the energy required for the transformation of the ice to vapor will involve the heat requirement in the following stages as:  

1) The energy required to melt ice = mass of ice × heat of fusion of water = 54g × 334 J/g  = 18036 J

  (because heat of fusion for water = 334 J/g)

2) The energy to heat water from 0 to 100 = mass of water × specific heat of water × change in temperature = 54g × 4.186 J/g°C × 100 °C = 22604.4 J  

lastly,

3) the energy required to vaporize 6.6g of water = mass of water × heat of vaporization of water = 6.6 × 2257 J/g = 14896.2 J  

Thus,

the total energy required to transform the ice cube to accomplish the transformation = 18036 + 22604.4 + 14896.2 = 55536.6 J.

Final answer:

To calculate gauge pressure in a tapered water pipe, apply Bernoulli's equation considering changes in radius, height, and flow rate.

Explanation:

The gauge pressure at a point in a tapered water pipe can be calculated using Bernoulli's equation. Given the information provided, we can determine the pressure difference between two points in the pipe based on the change in radius and height along the pipe.

In this scenario, the gauge pressure at a point 2.65 m downstream from the upper end can be calculated by considering the change in radius and height along the pipe as well as the flow rate of water through the pipe.

Understanding fluid dynamics and applying relevant equations helps determine the pressure variation in tapered pipes based on given parameters.

Answer:

The statue compress the stand by 3.68 x 10⁻⁵ m.

Explanation:

Change in length

            [tex]\Delta L=\frac{PL}{AY}[/tex]

Load, P = 3500 x 9.81 = 34335 N

Young's modulus, Y = 2.3 x 10¹⁰ N/m²

Area, A = 7.3 x 10⁻² m²

Length, L = 1.8 m

Substituting

         [tex]\Delta L=\frac{PL}{AE}=\frac{34335\times 1.8}{7.3\times 10^{-2}\times 2.3\times 10^{10}}=3.68\times 10^{-5}m[/tex]

The statue compress the stand by 3.68 x 10⁻⁵ m.

Answer:

B.Freezes

Explanation:

273 k=0 celcius which is converted to a melting point of ice which creates freezing point of water.

So clearly what I'm trying to say is that 273 K is when things turn into freezing ice.

Answer:

The engine's efficiency is 80%.

Explanation:

Given that,

Heat [tex]Q_{in}=75000\ J[/tex]

Remove heat [tex]Q_{out}=15000\ J[/tex]

We need to calculate the engine's efficiency

The efficiency is equal to the difference Input energy and output energy and divided by the input energy.

Using formula of efficiency

[tex]\eta=\dfrac{Q_{in}-Q_{out}}{Q_{in}}[/tex]

[tex]\eta=\dfrac{75000-15000}{75000}[/tex]

[tex]\eta\times100=80\%[/tex]

Hence, The engine's efficiency is 80%.

Explanation:

Light is clearly affected by gravity, just think about a black hole, but light supposedly has no mass and gravity only affects objects with mass. On the other hand, if light does have mass then doesn't mass become infinitely larger the closer to the speed of light an object travels.

Answer:

The correct answer is d volcanoes

Explanation:

Geothermal energy is a renewable energy that is obtained by harnessing heat from inside the earth that is transmitted through hot rock bodies or reservoirs by conduction and convection, where processes of interaction of groundwater and rocks occur. , giving rise to geothermal systems.

The geothermal energy of volcanoes allows to generate up to ten times more energy than an oil well. And it is an energy, more than renewable, eternal. In short, its characteristics are ideal for a greener future.

In Iceland, geothermal plants have already been put into operation, with an installed capacity of 665 MW, and are giving very good results

Iceland and New Zealand make extensive use of geothermal energy due to the numerous volcanoes in both countries. The heat originating from these volcanoes is harnessed to generate electricity. Their positions on or near the 'Ring of Fire' contribute to their high levels of volcanic activity.

Geothermal energy is used extensively in Iceland and New Zealand because both of these countries host multiple d. volcanoes. Volcanoes are locations where lava, which carries heat from within the Earth, rises to the surface. This heat can be harnessed to produce geothermal energy, which is transformed into electrical energy at geothermal energy plants, such as those found in Iceland and New Zealand.

Both countries are situated on or near the 'Ring of Fire', a belt of frequent volcanic and seismic activity that encircles the Pacific Ocean. This location contributes to their high volcanic activity. It's worth noting that the heat from volcanoes can serve as a source of renewable energy. For instance, the mantle 'hot spot' under the Hawaiian Islands has supplied the heat to maintain active volcanoes for at least 100 million years.

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